First, we prove that \(R/I\in \mathcal {B}_{\zeta }\) . Applying the functor \(\mbox{Hom}_{\sigma [R/I]}(R/I,-)\)  to the short exact sequence (REF ),
we have the following exact sequence
\(\mbox{Hom}_{\sigma [R/I]}(R/I,\mathsf {Tr}_{R/I}(Q_{0}))\rightarrow \mbox{Hom}_{\sigma [R/I]}(R/I,\mathsf {Tr}_{R/I}(Q_{1}))\rightarrow \mathsf {Ext}^{1}_{\sigma [R/I]}(R/I,T)\) 
From [1]}, we have isomorphisms
\(\mbox{Hom}_{R}(R/I,\mathsf {Tr}_{R/I}(Q_{i}))\cong \mbox{Hom}_{R}(R/I,Q_{i})\) , for \(i=1,2\) . Since \(\sigma [R/I]\)  is a full subcategory of \(\mathsf {Mod}\hspace{0.72229pt}R\) , we know that \( \mbox{Hom}_{\sigma [R/I]}(R/I,\mathsf {Tr}_{R/I}(Q_{0}))=\mbox{Hom}_{R}(R/I,\mathsf {Tr}_{R/I}(Q_{i}))\) , for \(i=1,2\) .
Note that \(\mathsf {Ext}^{1}_{\sigma [R/I]}(R/I,T)=0\) . Thus, we have that \(R/I\in \mathcal {B}_{\zeta }\) .
Assume that \(T\)  is not zero. It remains to show that \(T^{\mu }\)  is generated by \(R/I\)  and \(T^{\mu }\in \mathcal {B}_{\zeta }\)  for all sets \(\mu \)  since \(\sigma [R/I]\)  and \(\mathcal {B}_{\zeta }\)  are closed under submodules.
We divided the proof into several steps.
Step 1. We prove that \(\mathsf {Tr}_{R/I}(T^{\mu })\)  is a nonzero submodule of \(T^{\mu }\) .
If \(\mbox{Hom}_{R}(R/I,T )=0\) , then \(R/I\in \mathsf {Ker}\hspace{0.72229pt}\mbox{Hom}_{\sigma [R/I]}(-,T)\) . Note that \(R/I\in \mathcal {B}_{\zeta }\subseteq {^{\perp }T}\) . Then \(R/I\in \mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\)  since \(\mathsf {Ext}^{1}_{\sigma [R/I]}(R/I,T)\subseteq \mathsf {Ext}^{1}_{R}(R/I,T)\) . Since \(T\)  is a cotilting object in \(\sigma [R/I]\) , \(\mathsf {Cogen}\hspace{0.72229pt}_{\sigma [R/I]} T=\mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\) . Thus, \(R/I\)  is cogenerated by \(T\)  in \(\sigma [R/I]\) . Then there exist a monomorphism \(R/I\rightarrow \prod _{\theta \in \Theta }^{R/I}T_{\theta }\) , where \(\prod _{\theta \in \Theta }^{R/I}T_{\theta }\)  is the product of \(T\)  in \(\sigma [R/I]\) . By [2]}, there exists a monomorphism \(\prod _{\theta \in \Theta }^{R/I}T_{\theta }\rightarrow T^{\Theta }\) , where \(T^{\Theta }\)  is the product of \(T\)  in \(\mathsf {Mod}\hspace{0.72229pt}R\) . Thus, there is a monomorphism \(R/I\rightarrow T^{\Theta }\) . It implies that \(\mbox{Hom}_{R}(R/I,R/I)=0\) . Thus, \(R/I=0\)  and so \(\mathsf {Tr}_{R/I}(Q_{0})=0\) . Form the exact sequence (REF ), we know that \(T=0\) . It is a contradiction. Hence, \(\mbox{Hom}_{R}(R/I,T )\ne 0\)  and so \(\mbox{Hom}_{R}(R/I,T^{\mu } )\ne 0\) . Therefore, \(\mathsf {Tr}_{R/I}(T^{\mu })\)  is a nonzero module.
Step 2. We prove that \(T^{\mu }\)  is generated by \(R/I\) .
From [1]}, we have an isomorphism
\(\mbox{Hom}_{R}(R/I, \mathsf {Tr}_{R/I}(T^{\mu }))\cong \mbox{Hom}_{R}(R/I, T^{\mu }).\) 
Since \(\mathsf {Tr}_{R/I}(T^{\mu })\)  is generated by \(R/I\) , there exists a nonzero epimorphism of \(R\) -modules \(f:{R/I}^{(\lambda )} \rightarrow \mathsf {Tr}_{R/I}(T^{\mu })\)  where \(\lambda \)  is a set. For the set \(\lambda \) , there exists isomorphism
\(\mbox{Hom}_{R}({R/I}^{(\lambda )}, \mathsf {Tr}_{R/I}(T^{\mu }))\cong \mbox{Hom}_{R}({R/I}^{(\lambda )}, T^{\mu }).\) 
Then there exists an epimorphism \(f^{\prime }:{R/I}^{(\lambda )} \rightarrow T^{\mu }\) . Thus, \(T^{\mu }\)  is generated by \(R/I\) . Thus, \(\mathsf {Cogen}\hspace{0.72229pt}T\subseteq \sigma [R/I]\)  since \(\sigma [R/I]\)  is closed under submodules.
Step 3. We claim that \(\mbox{Hom}_{R}(T^{\mu },M/\mathsf {Tr}_{R/I}(M))=0\)  for any \(M\in \mathsf {Mod}\hspace{0.72229pt}R\) .
We set \(h\in \mbox{Hom}_{R}(T^{\mu },M/\mathsf {Tr}_{R/I}(M))\) . For the epimorphism \(f^{\prime }:{R/I}^{(\lambda )} \rightarrow T^{\mu }\) , let \(\varepsilon _{\lambda _{i}}:(R/I)_{\lambda _{i}}\rightarrow {R/I}^{(\lambda )}\)  be the \(\lambda _{i}\) -th canonical embedding with \(\lambda _{i}\in \lambda \) . Then for any \(t\in T^{\mu }\) , there exists a family \(\lbrace \overline{r_{i}}\rbrace ^{n}_{i=1}\in R/I\)  such that \(\sum _{i}^{n}f^{\prime }\varepsilon _{\lambda _{i}}(\overline{r_{i}})=t\) . Hence, \(h(t)=\sum _{i}^{n}hf^{\prime }\varepsilon _{\lambda _{i}}(\overline{r_{i}})\in \mathsf {Tr}_{R/I}(M/\mathsf {Tr}_{R/I}(M))\) . Note that \(\mathsf {Tr}_{R/I}(M/\mathsf {Tr}_{R/I}(M))=\mathsf {Rej}_{I}(M/\mathsf {Rej}_{I}(M))=0\) . Thus, we know that \(\mbox{Hom}_{R}(T^{\mu },M/\mathsf {Tr}_{R/I}(M))=0\) .
Step 4. We prove that \(T^{\mu }\in \mathcal {B}_{\zeta }\) .
Since \(T^{\mu }\)  is generated by \(R/I\) , \(T^{\mu }\in \sigma [R/I]\) . By Lemma REF , \(T^{\mu }=\prod _{\kappa \in \mu }^{R/I}T_{\kappa }\)  in \(\sigma [R/I]\) . Applying the functor \(\mbox{Hom}_{\sigma [R/I]}(T^{\mu },-)\)  to the short exact sequence (REF ), we have the following exact sequence
\(\mbox{Hom}_{\sigma [R/I]}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{0}))\rightarrow \mbox{Hom}_{\sigma [R/I]}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{1}))\rightarrow \mathsf {Ext}^{1}_{\sigma [R/I]}(T^{\mu },T)\) 
Since \(T\)  is a partial cotilting object in \(\sigma [R/I]\)  and \(\sigma [R/I]\)  is a full subcategory of \(\mathsf {Mod}\hspace{0.72229pt}R\) , we have \(\mathsf {Ext}^{1}_{\sigma [R/I]}(T^{\mu },T)=\mathsf {Ext}^{1}_{\sigma [R/I]}(\prod _{\kappa \in \mu }^{R/I}T_{\kappa },T)=0\)  and so,
\(\mbox{Hom}_{R}(T^{\mu },\mathsf {Tr}_{R/I}(\zeta )):\mbox{Hom}_{R}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{0}))\rightarrow \mbox{Hom}_{R}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{1}))\) 
is surjective. Applying \(\mbox{Hom}_{R}(T^{\mu },-)\)  to the short exact sequence (REF ), we have the following exact sequence
\(\mbox{Hom}_{R}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{0}))\rightarrow \mbox{Hom}_{R}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{1}))\rightarrow \mathsf {Ext}^{1}_{R}(T^{\mu },T)\rightarrow \mathsf {Ext}^{1}_{R}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{0})).\) 
Now, we shall prove that \(\mathsf {Ext}^{1}_{R}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{0}))=0\) .
Applying \(\mbox{Hom}_{R}(T,-)\)  to the short exact sequence
\(0\rightarrow \mathsf {Tr}_{R/I}(Q_{0})\rightarrow Q_{0}\rightarrow Q_{0}/\mathsf {Tr}_{R/I}(Q_{0})\rightarrow 0,\) 
we have the following exact sequence
\(\mbox{Hom}_{R}(T^{\mu },Q_{0}/\mathsf {Tr}_{R/I}(Q_{0}))\rightarrow \mathsf {Ext}^{1}_{R}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{0}))\rightarrow \mathsf {Ext}^{1}_{R}(T^{\mu },Q_{0}).\) 
From the injectivity of \(Q_{0}\)  in \(\mathsf {Mod}\hspace{0.72229pt}R\)  and Step 3, we have that \(\mathsf {Ext}^{1}_{R}(T^{\mu },\mathsf {Tr}_{R/I}(Q_{0}))=0\) .
From the exact sequence (REF ), we obtain that \(\mathsf {Ext}^{1}_{R}(T^{\mu },T)=0\) . That is, \(T^{\mu }\in {^{\perp }T}\) .
Now, we have a short exact sequence \(0\rightarrow K\rightarrow {R/I}^{(\lambda )}\xrightarrow{} T^{\mu }\rightarrow 0\)  with \({R/I}^{(\lambda )}\in \mathcal {B}_{\zeta }\) . Since \( \mathcal {B}_{\zeta }\)  is closed under submodules, \(K\in \mathcal {B}_{\zeta }\) . Then we have the following exact commutative diagram
\({&0[d]&0[d]&0[d]&\\0[r]&\mbox{Hom}_{R}(T^{\mu }, T)[r][d]&\mbox{Hom}_{R}(T^{\mu }, Q_{0})[r][d]&\mbox{Hom}_{R}(T^{\mu }, Q_{1})[d]&\\0[r]&\mbox{Hom}_{R}({R/I}^{(\lambda )}, T)[r][d]&\mbox{Hom}_{R}({R/I}^{(\lambda )}, Q_{0})[r][d]&\mbox{Hom}_{R}({R/I}^{(\lambda )}, Q_{1})[r][d]&0\\0[r]&\mbox{Hom}_{R}(K, T)[d][r]&\mbox{Hom}_{R}(K, Q_{0})[r][d]&\mbox{Hom}_{R}(K, Q_{1})[r][d]&0\\ &0&0&0& }\) 
Applying Snake Lemma, we know that the first row is a short exact sequence, and so \(T^{\mu }\in \mathcal {B}_{\zeta }\) .
